4 numbers with 3 + or x operations and parenthese 
I tried list out the possibilities but there seem to be too many after a while. Is there a systematic way to count this? Do the value of the numbers matter? Thanks!
 A: I would start by listing the ways to parenthesize the operations.  You can think of it as the order of doing the operations, of which there are $3!=6$, but $132$ and $231$ will always give the same answer because if you do the middle one last it doesn't matter what order you do the end ones.
There are $2^3=8$ ways to fill the operations into the boxes.   That gives $40$ possibilities so far.  
Then note that both operations are commutative and associative.  You can prune the list by noting that if all the signs are plus, it doesn't matter how you parenthesize.  Some of the other combinations will be demonstrably equal as well.  
There are $4!$ ways to fill in the blanks, but many of them are again demonstrably equal.  For example, if you are doing $(\_ + \_)\times (\_+\_)$ you can interchange the pairs in parentheses and also the order of the pairs.  There are only three different ways to fill this one in.  
It matters what numbers you are given because there might be "accidental" equalities.  You could use $a,b,c,d$ for the numbers and come up with the list of possible results, but if $c=1$ then $abc+d$ is equal to $ab+cd$ and the number of possible results will be reduced.  One way to make sure you have the maximum is to have all four numbers be primes of such different magnitude that no products or sums can match. 
It is a lot of work, no matter how you do it.
